If X follows a binomial distribution with parameters `n=6` and p. If `4(P(X=4))=P(X=2)` , then `P=`

Options

A. `1//2`
B. `1//4`
C. `1//6`
D. `1//3`

Answer

To solve the problem, we need to find the value of \( p \) given that \( X \) follows a binomial distribution with parameters \( n = 6 \) and \( p \), and the condition \( 4P(X=4) = P(X=2) \).

### Step-by-Step Solution:

1. **Understand the Binomial Probability Formula**:
The probability mass function for a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]
where \( n \) is the number of trials, \( r \) is the number of successes, and \( p \) is the probability of success on each trial.

2. **Set Up the Equations**:
We need to calculate \( P(X = 4) \) and \( P(X = 2) \) using the binomial formula:
- For \( P(X = 4) \):
\[ P(X = 4) = \binom{6}{4} p^4 (1-p)^{2} \]
- For \( P(X = 2) \):
\[ P(X = 2) = \binom{6}{2} p^2 (1-p)^{4} \]

3. **Substitute into the Given Condition**:
According to the problem, we have:
\[ 4P(X = 4) = P(X = 2) \]
Substituting the expressions we derived:
\[ 4 \left( \binom{6}{4} p^4 (1-p)^{2} \right) = \binom{6}{2} p^2 (1-p)^{4} \]

4. **Calculate Binomial Coefficients**:
The binomial coefficients are:
\[ \binom{6}{4} = \binom{6}{2} = 15 \]
So, we can simplify the equation:
\[ 4 \cdot 15 p^4 (1-p)^{2} = 15 p^2 (1-p)^{4} \]
Dividing both sides by 15 gives:
\[ 4 p^4 (1-p)^{2} = p^2 (1-p)^{4} \]

5. **Rearranging the Equation**:
We can rearrange this equation:
\[ 4 p^4 (1-p)^{2} - p^2 (1-p)^{4} = 0 \]
Factor out \( p^2 (1-p)^{2} \):
\[ p^2 (1-p)^{2} (4p^2 - (1-p)^{2}) = 0 \]

6. **Solve the Factors**:
The first factor \( p^2 = 0 \) gives \( p = 0 \) (not valid since \( p \) must be between 0 and 1). Now, solve the second factor:
\[ 4p^2 - (1 - 2p + p^2) = 0 \]
Simplifying gives:
\[ 4p^2 - 1 + 2p - p^2 = 0 \implies 3p^2 + 2p - 1 = 0 \]

7. **Use the Quadratic Formula**:
We can apply the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = 2, c = -1 \):
\[ p = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6} \]

8. **Calculate the Roots**:
- First root:
\[ p = \frac{2}{6} = \frac{1}{3} \]
- Second root:
\[ p = \frac{-6}{6} = -1 \quad (\text{not valid}) \]

9. **Conclusion**:
The valid solution is:
\[ p = \frac{1}{3} \]

### Final Answer:
\[ P = \frac{1}{3} \]